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Le document semble contenir des informations sur une transcription musicale, probablement d'une ?uvre musicale, comprenant divers ¨¦l¨¦ments tels que des indications de tonalit¨¦ et des structures musicales. Les mentions de trombones et les noms de morceaux sugg¨¨rent qu'il est destin¨¦ ¨¤ des musiciens ou ¨¤ des performances. Toutefois, le texte est en grande partie illisible ou corrompu, rendant difficile une compr¨¦hension pr¨¦cise de son contenu.????????????? ?????????
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AMOR A MEDIO TIEMPO(BOBBY VALENTIN)
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Transcripcion? Marcos Burbano
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Sud?tingesn?s trupmenin?s lygtys
Sud?tingesn?s trupmenin?s lygtysLina ?a
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This document contains solutions to three equations involving fractions equal to values.
The first equation, x^2 + x - 2 = 0, is solved to find the values of x that satisfy the equation, which are x = 1 and x = -2.
The second equation, (x-1)/(x+1) = 3, is solved to find the value of x that makes the fraction equal to 3, which is x = -2.
The third equation, x/(x-1) - 1/(x+2) - 3/(x-1)(x+2) = 0, is solved but does not have any real number solutions for x.Ad
?Aknys
- 1. ?aknys
- 2. Neneigiamasis skai?ius b ¡Ý 0, kuriam yra teisinga lygyb? b n = a , vadinamas aritmetine n -ojo laipsnio ?aknimi i? a . a ¡Ý 0, o n ¡Ý 2 ¨C lyginis nat¨±ralusis skai?ius. ?ymime b = a; n a ¨C n -ojo laipsnio ?aknies po?aknis, b ¨C n -ojo laipsnio ?aknies reik?m?.
- 3. Pvz.: 16 = 4, (n = 2), nes 4 2 = 16 ir 4 > 0; 4 6 81 = 3, (n = 4), nes 34 = 81 ir 3 > 0; 64 = 2, (n = 6), nes 26 = 64 ir 2 > 0. Lyginio laipsnio ?aknis su neigiamuoju po?akniu neturi prasm?s. Pvz.: 4 - 16 , ? 3
- 4. Sakykime a ¨C bet koks realusis skai?ius, n ¡Ý 3 ¨C nelyginis nat¨±ralusis skai?ius. Skai?ius b (jo ?enklas sutampa su a ?enklu), kuriam yra teisinga lygyb? b n = a , vadinamas n -ojo laipsnio ?aknimi i? a. Pvz.: 3 3 5 8 = 2, (n = 3), nes 23 = 8; ? 8 = ?2, (n = 3), nes (?2) = ?8; 3 32 = 2, (n = 5), nes 25 = 32;
- 5. Kai ?aknies laipsnio rodiklis nelyginis, yra teisinga lygyb? n Pvz.: 3 5 ? a = ?n a ? 2 = ?3 2 ; ? 32 = ? 32 = ?2. 5
- 6. n-ojo laipsnio ?aknys i? neigiam?j? skai?i?: n a =a n Pvz.: n kai n ¨C lyginis. (?5) = ? 5 = 5; 2 a =a n 3 4 (?5) = ? 5 = 5; 4 kai n ¨C nelyginis. (?5) = ?5; 3 5 (?5) = ?5; 5
- 7. ?akn? savyb?s n Pvz. : 3 a ? b = a ?b n n 2 ? 3 500 = 3 2 ? 500 = 3 1000 = 3 103 = 10
- 8. n Pvz. : 4 n a n a a: b= n = b b n 3 4 3 3 4 3 16 4 3:4 = : = ? = 16 = 4 2 4 = 2 16 1 16 1 3
- 9. ( a) n Pvz. : k ( 4) = a n 3 k = 4 = 64 = 8 3
- 10. a = n k Pvz. : 3 4 3 2= n?k 3?4 a 2= 2 12 7 = 2?3 7 = 6 7
- 11. n?k Pvz. : 6 8 a = a k n 3 5 = 2?3 5 = 5 16 = 2?4 2 = 2 3 4
- 12. a b = a ?b n n Pvz. : n 4 3 = 4 ? 3 = 16 ? 3 = 48 2 2 5 = 2 ? 5 = 8 ? 5 = 40 3 3 3